How to write exact 1MB array in C? - c

Question

I want to initialize an array of size 1MB. So my goal is finally write that 1MB to a file.
I am curious every time i use this formula it is giving less than 1mb.
int len = (1048576)/sizeof(int);
data = (int *) malloc(len);
What is correct way ?
Thank you
Edit - As per the comments I have changed the code .
#include <stdio.h>
#include <stdlib.h>
#include <mpi.h>
#include <string.h>
int main(int argc, char *argv[]) {
int *data;
int bytes = (1024*1024);
data = (int *) malloc(bytes);
for(int i=0;i<bytes;i++){
data[i] = (int)rand();
printf("%d",data[i]);
}
return 0;
}
After compiling it and I tried dumping the data like below
mpicc -o a mpiFileSize.c
./a > dump.dat
Now I see the file size of dump.dat. Why its 2.5MB ?

Answer 1

Try this
#include <stdio.h>
#include <stdlib.h>
#include <mpi.h>
#include <string.h>
int main(int argc, char *argv[]) {
char *data;
int bytes = (1024*1024);
data = (char *) malloc(bytes);
for(int i=0;i<bytes;i++){
data[i] = (char) rand();
printf("%c",data[i]);
}
return 0;
}
You shoul use character instead of integer.

Answer 2

Although it was already properly answered.
Just a plus to the answer, if one wants to choose the amount of MBs to allocate would make something like:
#include <malloc.h>
#define Mebabyte (1024 * 1024)
int main(int argc, char** argv)
{
void* data = malloc(2 * Megabyte);
// Do your work here...
free(data);
return 0;
}
If you wanted to allocate more than 2 MBs just change the 2.
As already stated before do not use integers as it's going to have more than 1 byte of size. Instead use char or unsigned char. And as stated by another post, there's no need to cast the result of malloc since void* can be turned to a pointer to any type (and in fact it's done implicitly by the compiler).
see: Why does this code segfault on 64-bit architecture but work fine on 32-bit?